
In this formula b = 6
a = (2 x 20^4) / b^6
So the answer is 6.8587
If I already knew that a = 6.8587, and b was the unknown, how would I solve that problem so that the answer to b didn't come out in exponential form. I mean, so that the answer would come out as "6" instead of something to the something power?
If it did come out as something to the something power, how would I convert that to a regular number?

Holy cow...your asking me to remember my pre calculus days. When your trying to solve to that power things get tuff. Well, I will tell you right now Im not going to do it!
Im glad thats over.
To put the world in order, we must first put the nation in order; to put the nation in order, we must put the family in order; to put the family in order, we must cultivate our personal life; and to cultivate our personal life, we must first set our hearts right.
 Confucius

Mid
You are going to sprain your brain if you keep this up.

midhvac,
Do the basic algebra and up end up with:
b=(2x20^4/a)all raised to the 1/6
You need a scientific calculator to get what b is. That is, 46656.07 ^1/6
plug in all the numbers and the answer is 6.
Is this what you were asking?

Originally posted by midhvac
In this formula b = 6
a = (2 x 20^4) / b^6
So the answer is 6.8587
Perhaps some number theory may be appropriate here.
'a' in this example is equal to exactly 320,000 / 46,656 or approximately 6.85871056241426611796982167352538 (using Microsoft calculator )
'a' is considered a 'rational' number because it can be wriiten as a fraction, i.e., x / y where x and y are integers.
Originally posted by midhvac
If I already knew that a = 6.8587, and b was the unknown, how would I solve that problem so that the answer to b didn't come out in exponential form. I mean, so that the answer would come out as "6" instead of something to the something power?
Ok, lets assume 'a' = 7, what is 'b'?
b = (320,000 / 7)^(1/6)
where 'b' is approximately 5.979643919487405903197485402773 (again using Microsoft calculator)
In this case, 'b' is an irrational number. There is no fraction, x / y, that can define 'b' where x and y are integers. Think about it! No integer numbers whatsoever. Raising this expression to the 1/6th power causes this to occur.
Mmmmm... so does this help?

You guys rock! Thanks! I've got a scientific calculator and the 1/6 power works.

Originally posted by condenseddave
Mid
You are going to sprain your brain if you keep this up.
Nah, I put an elastic support around my head, just in case.

Hey! Wait a minute! Why are we using 1/6? Is it because the answer to b is 6? So I have to know the answer before I can work the problem?

Originally posted by midhvac
Originally posted by condenseddave
Mid
You are going to sprain your brain if you keep this up.
Nah, I put an elastic support around my head, just in case.
Let's put that elastic support to work...
Using your calculator, solve the following:
x^3 = x * x * x = 27
or x = (27)^(1/3)
A little figuring will show that 'x' must equal 3, i.e., 3 * 3 * 3 = 27
Why does the calculator fail to figure this out?

Originally posted by midhvac
Hey! Wait a minute! Why are we using 1/6? Is it because the answer to b is 6? So I have to know the answer before I can work the problem?
Nope. Using apropriate math:
a = (2 x 20^4) / b^6
a = 320,000 / b^6
b^6 = 320,000 / a
(b^6)^(1/6) = (320,000 / a)^(1/6)
b = (320,000 / a)^(1/6)

I think I am going into another field. I will never figure this stuff out!!!

You big bunch of math nerds.

Originally posted by mattm
You big bunch of math nerds.
Yea , bet they got pocket protectors too>)
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